JEE 2010..thermo

One mole of an ideal gas in initial state A undergoes a cyclic process ABCA as shown. Its pressure at A is Po. then

(a) Internal energies at A and B are same
(b) Work done by the gas in the process AB is PoVo ln4
(c) pressure at C is Po/4
(d) Temperature at C is To/4

shouldn't the answer be (a),(b),(c),(d)??? fiitjee says only a and b

18 Answers

39
Pritish Chakraborty ·

Why exactly won't the line pass through origin??
It's ideal conditions. PV = nRT.

If we plot a graph between V and T(obviously the other quantities are constant), we get a straight line of the form y = mx where m = nR/P...
Doesn't that pass through the origin???

If you think P isn't constant, that can't be. Had that been the case, we'd get a many sloped curve, not a straight line.
(Discussed this with govind. Any corrections are welcome)

1
Shahbaz Ali ·

IIT PROFESSORS SAID THAT THOSE WHO WILL MARK ALL 4 CHOICES WILL BE CAUGHT BCOZ THEY THINK THAT THOSE BCOZ OF THAT PARTIAL MARKS TECHNIQUE THEY WILL MARK ALL 4 WHICH IS ABSOLUTELY ABSURD THIS QUESTION HAD REALL ABCD AS THE SOLUTION

1
Akshay Pamnani ·

when did they say this??

1
RAY ·

IIT SAYS NO ANSWER CAN BE ABCD!! :D

1
jabbujabbu ·

Option No. (A) and (B) ,here , seems easy to go through .

Regarding the (C) and (D) options ,there is a controversy going on about whether the line BC passes through origin or not.

Now,

Remember Charles Law - " At constant pressure , the volume of a given mass of gas is directly proportional to the absolute temperature of the gas " .

Another way to state the same is like - " Pressure remaining constant , the volume of a given mass of gas increases or decreases by 1/273 part of it's volume at 0°C for a rise or fall in temperature of 1°C "

i.e. volume of gas at t°C , V = V0(1 + t/273) , where V0 is the volume of the gas at 0°C .

Considering this expresion let us find the volume of gas at -273 °C i.e 0 K ,

V= V0(1 - 273/273) = 0

This implies , w.r.t the question , that the line BC (V/T = constant) must pass through the origin . Again, it is well known that ideal gases have zero volume at 0 K.

So, why aren't the coaching institutes paying a real attention to this fact ?
Do they have the doubt that whether the temperature on the x-axis is in °Centigrade or Kelvin ?

Obviously , it's in Kelvin , Ideal gas laws are valid for Kelvin scale only !

Also what Pritish and Subhomoy said earlier is correct and the same.

Hence i think the answers should be (A)(B)C)(D).

1
Akshay Pamnani ·

IIT would give anwers jab mamla thanda ho jayega
jaisa sab bhool jaye tab tak

1
RAY ·

does iit give soltuions ?? ?

39
Pritish Chakraborty ·

We'll have to wait for IIT's own solutions..

49
Subhomoy Bakshi ·

ideal gases have 0 volume at 0 absolute temperature!!

so no y intercept!! since (0,0) is a point on graph!!

so

answer should be no way be a and b

correct set of answers must be a b c d

1
student ·

very easy one
typical iit kind
A, B
is the asnwer

in the other procees neither slope is given nor any relation nor any cordinate

so unknown process are happneing

1
RAY ·

WTF!!! did they mention in the optics question whether the line was passing out straight of the lens or not???wht did u do there????start writing coordinates.....dint u use ur eye estimation or a ruler to chek dat!!!

@ rajat : no offence....but if u can answer my query...how did u find the refractive index relations in the second paper match the column!!!

21
amit sahoo ·

i also marked ABCD in the exam but now i think it should be AB only.

1
rajatjain_ix ·

hey i also SAY THAT ANS IS A,B ,,,,,,, SINCE CD LINE MAY OR MAY NOT BE PASSING THROUGH ORIGIN....!!

1
The Enlightened One - jsg ·

arrey ans is : a,b,c,d

1
student ·

THIS WAS THE KIND OF QUESTION THAT EVERY QUESTION SHOULD HAVE BEEN

1
Akshay Pamnani ·

dekhna answers hone
AB or ABCD

1
Shreyan ·

ohhhhh...okkk IIT is clever, eh??? [3]

33
Abhishek Priyam ·

These are the questions IIT keep in the paper to make the ranking or marks unpredictable.. in these they can do anything...

There has been a case against the IITJEE selection procedure to make it more transparent. Lets see...

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